# JEE Main & Advanced Mathematics Trigonometric Equations Ex-central Triangle

## Ex-central Triangle

Category : JEE Main & Advanced

Let ABC be a triangle and I be the centre of incircle. Let ${{I}_{1}}$, ${{I}_{2}}$ and ${{I}_{3}}$ be the centres of the escribed circles which are opposite to A, B, C respectively then ${{I}_{1}}{{I}_{2}}{{I}_{3}}$ is called the Ex-central triangle of $\Delta ABC$.

${{I}_{1}}{{I}_{2}}{{I}_{3}}$ is a triangle, thus the triangle ABC is the pedal triangle of its ex-central triangle ${{I}_{1}}{{I}_{2}}{{I}_{3}}$. The angles of ex-central triangle ${{I}_{1}}{{I}_{2}}{{I}_{3}}$ are  ${{90}^{o}}-\frac{A}{2},\,\,{{90}^{o}}-\frac{B}{2},\,\,{{90}^{o}}-\frac{C}{2}$ and sides are ${{I}_{1}}{{I}_{3}}=4R\cos \frac{B}{2};\,\,{{I}_{1}}{{I}_{2}}=4R\cos \frac{C}{2};\,\,{{I}_{2}}{{I}_{3}}=4R\cos \frac{A}{2}$.

Area and circum-radius of the ex-central triangle

Area of triangle

$=\frac{1}{2}$ (Product of two sides) $\times$ (sine of included angles)

$\Delta =\frac{1}{2}\,\,\left( 4R\cos \frac{B}{2} \right)\,\,.\,\,\left( 4R\cos \frac{C}{2} \right)\times \sin \left( {{90}^{o}}-\frac{A}{2} \right)$

$\Delta =8{{R}^{2}}\cos \frac{A}{2}.\cos \frac{B}{2}.\cos \frac{C}{2}$

Circum-radius $=\frac{{{I}_{2}}{{I}_{3}}}{2\sin {{I}_{2}}{{I}_{1}}{{I}_{3}}}=\frac{4R\cos \frac{A}{2}}{2\sin \left( {{90}^{o}}-\frac{A}{2} \right)}=2R$.

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